Our aim is to find a condition for the existence of a hyperbolic
through the vertices of a hyperbolic triangle in terms of the lengths
of its sides.
We begin with a theorem which holds in both euclidean and hyperbolic
geometry, i.e. a theorem of neutral geometry.
In the theorem, we consider the size of angles, and assume that each
angle is taken in the range (0,π). That is to say, we take <XYZ as the
measure of the internal angle at vertex Y in ΔXYZ.
the angle theorem
For any triangle ΔABC, and any point P which is not on any
of the lines AB, BC, CA. Let <APB=γ, <BPC=α, <CPA=β, then
cos2(α) + cos2(β) + cos2(γ) =
2cos(α)cos(β)cos(γ) + 1.
Although this result has nice applications in euclidean geometry, we
concentrate on the hyperbolic case.
Suppose that ΔABC is a hyperbolic triangle
and that P is a point
not on any of the hyperbolic lines AB, BC, CA.
Let d(A,B)=c, d(B,C)=a, d(C,A)=b, d(P,A)=r, d(P,B)=s, d(P,C)=t.
Then G(cosh(r),cosh(s),cosh(t),cosh(a),cosh(b),cosh(c)) = 0,
where G(R,S,T,X,Y,Z) is given by
-2(RS-Z)(ST-X)(TR-Y) - (R2-1)(S2-1)(T2-1).
This is obtained from the angle theorem and the cosine rule
for the angles at P.
It is easy to see that G is quadratic in each of R, S, T.
A calculation shows that the coefficient of T2 is Z2-1.
Since we will have Z = cosh(c) > 1, G is quadratic in T.
Now observe that, if ΔABC has a hyperbolic circumcircle K(P,r), then
we can apply the lemma to the hyperbolic centre P. We then have
R = S = T = cosh(r), and the condition becomes
(R2-1)3 = 0,
where X =cosh(a), Y = cosh(b), Z = cosh(c), R = cosh(r).
The left hand side is clearly a polynomial in R2, and appears to involve
terms in R4 and R6. If we multiply out and simplify, we get the equation
F(cosh(a),cosh(b),cosh(c))R2 = Δ2(cosh(a),cosh(b),cosh(c)),
F(X,Y,Z) = 3+2(XY+YZ+ZX)-2(X+Y+Z)-(X2+Y2+Z2), and
Δ2(X,Y,Z) = 1+2XYZ-(X2+Y2+Z2).
This is the the circumradius equation
By expanding then simplifying we get the difference formula
Δ2(X,Y,Z)-F(X,Y,Z) = 2(X-1)Y-1)(Z-1).
We have met Δ2 before, in connection with the trigonometry of hyperbolic
triangles. From that context, we know that it must be positive. We take
as the positive root of this quantity.
It follows that, as R2 > 0, we must have F(cosh(a),cosh(b),cosh(c)) > 0.
We have arrived at the following
necessary condition for a hyperbolic circumcircle
If the hyperbolic triangle ΔABC has a hyperbolic circumcircle, then
(1) F(cosh(a),cosh(b),cosh(c)) > 0, and
(2) the hyperbolic radius r is determined by
cosh(r) = Δ(cosh(a),cosh(b),cosh(c))/F½(cosh(a),cosh(b),cosh(c)).
The other hyperbolic functions of r are easily determined from cosh(r).
The manipulation is a little tedious, but we obtain
sinh2(r) = 2(cosh(a)-1)cosh(b)-1)(cosh(c)-1)/F(cosh(a),cosh(b),cosh(c)),
tanh2(r) = 2(cosh(a)-1)cosh(b)-1)(cosh(c)-1)/Δ2(cosh(a),cosh(b),cosh(c)).
The formula for sinh(r) shows again that we must have F > 0.
The formula for tanh(r) gives an alternative formulation of the condition
Alternative necessary condition
If ΔABC has a hyperbolic circumcircle, then we must have
2(cosh(a)-1)cosh(b)-1)(cosh(c)-1) < Δ2(cosh(a),cosh(b),cosh(c)).
This is immediate as tanh(r) < 1.
If we observe that cosh(x)-1 = 2sinh2(½x), the formula for tanh(r) yields
tanh(r) = 4sinh(½a)sinh(½b)sinh(½c)/Δ(cosh(a),cosh(b),cosh(c))
The removal of the need to take a root is an illusion - Δ involves a root.
There are various ways to rewrite the formulae for r. Some are given as
Our main purpose is to show that either of our conditions are also sufficient
to guarantee the existence of a circumcircle. This is not so easy. We give a
proof in a separate page.